Maharashtra Board SSC Class 10 Maths 1 Question Paper 2024 with Solutions Answers Pdf Download.

## SSC Maths 1 Question Paper 2024 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours

Max. Marks : 40

General Instructions:

- All questions are compulsory.
- Use of a calculator is not allowed.
- The numbers to the right of the questions indicate full marks.
- In case of MCQs [Q. No. 1(A)] only the first attempt will be evaluated and will be given credit.

Question 1.

(A) Choose the correct alternative from given : [4]

(i) If 3 is one of the root of the quadratic equation kx^{2} – 7x +12 = 0, then k =

(A) 1

(B) – 1

(C) 3

(D) -3

Answer:

(A) 1

Explanation:

One of the root of equation kx^{2} – 7x + 12 = 0 is 3

∴ Now let x = 3 in the equation,

kx^{2} – 7x + 12 = 0

9k – 21 + 12 = 0

9k – 9 = 0

9k = 9

k = 1

(ii) To draw the graph of x + 2y = 4, find x when y = 1:

(A) 1

(B) 2

(C) -2

(D) 6

Answer:

(B) 2

Explanation:

Given, equation,

x + 2y = 4

When y = 1

∴ x + 2 × 1 = 4

x = 4 – 2

x = 2

(iii) For an A.P., t_{7} = 4, d = – 4, then a = ___.

(A) 6

(B) 7

(C) 20

(D) 28

Answer:

(D) 28

Explanation:

Given,

t_{7} = 4, d = -4

t_{7} = 4

a + (n – 1)d = 4

a + (7 – 1) (- 4) = 4

a + 6(- 4) = 4

a – 24 = 4

a = 28

(iv) In the format of GSTIN, there are _____ alpha-numerals.

(A) 9

(B) 10

(C) 15

(D) 16

Answer:

(C) 15

Explanation : GSTIN has 15 alpha numerals.

(B) Solve the following subquestions: [4]

(i) If 17x + l5y =11 and 15x + 17y = 21, then find the value of x – y.

Solution:

17x + 15y = 11 …… (1)

15x + 17y = 21 ….. (2)

Subtract equation (2) from (1),

Divide by 2,

x – y = -5.

(ii) Find first term of the sequence t_{n} = 3n – 2.

Solution:

t_{n} = 3n – 2

Put n = 1

t_{1} = 3(1) – 2

t_{1} = 3 – 2

t_{1} = 1.

(iii) If the face value of a share is ₹ 100 and market value is ₹150. If rate of brokerage is 2%, find brokerage paid on one share.

Solution:

FV = ₹100

MV = ₹150

Rate of brokerage = 2% of MV

Brokerage paid on one share = 2% of MV

= \(\frac{2}{100}\) × 150 = \(\frac{30}{10}\) = ₹3

Brokerage paid on one share = ₹3.

(iv) Two digit numbers are formed using digits 2,3 and 5 without repeating a digit. Write the sample space.

Solution:

Let ‘S’ be the sample space

S = {23, 25, 32, 35, 52, 53}

n(S) = 6.

Question 2.

(A) Complete the following activities and rewrite it (any two): [4]

(i) If (0,2) is the solution of 2x + 3y = k, then to find the value of k, complete the following activity:

Activity:

(0,2) is the solution of the equation 2x + 3y = k.

Solution:

Activity:

(0, 2) is the solution of the equation 2x + 3y = k

Put x = __o__ and y = __2__ in the given equation;

2 × __0__ + 3 × 2 = k

0 + 6 = k

k = __6__.

(ii) If 2 and 5 are the roots of the quadratic equation, then complete the following activity to form quadratic equation:

Activity:

Let α = 2 and β = 5 are the roots of the quadratic equation. Then quadratic equation is:

Solution:

Activity:

Let α = 2 and β = 5 are the roots of the quadratic equation. Then quadratic equation is:

x^{2} – (α + β)x + αβ = 0

x^{2} – (2 + __5__)x + __2__ × 5 = 0

x^{2} – __7__x + __10__ = 0.

(iii) Two coins are tossed simultaneously. Complete the following activity to write the sample space and the given events A and B in the set form :

Event A : To get at least one head.

Event B : To get no head.

Activitiy:

Two coins are tossed simultaneously.

Solution:

Activity:

Two coins are tossed simultaneously.

∴ Sample space, S = {__TT__, HT, TH, __HH__}

Event A : To get at least one head.

A = {__HH__, HT, TH}

Event B : To get no head.

B = {TT}.

(B) Solve the following subquestions (any four): [8]

(i) □ABCD is a rectangle. Write two simultaneous equations using information given below in the diagram, in the form of ax + by = c:

Solution:

In □DABCD,

Side AB = Side DC [opposite sides of rectangle]

Side AD = Side BC

2x + y + 8 = 4x – y

4x – 2x – y – y – 8 = 0

2x – 2y = 8

Divide by 2,

x – y = 4 ……. (1)

Similarly, 2y = x + 4

x – 2y = -4 …….. (2)

x – y = 4 and x – 2y = -4 are the required simultaneous equations.

(ii) Solve the following quadratic equation using factorisation method :

x^{2} + x – 20 = 0.

Solution:

x^{2} + x – 20 =0

x^{2} + 5x – 4x – 20 = 0

x(x + 5) – 4(x + 5) = 0

(x + 5)(x – 4) = 0

(x + 5) = 0 or (x – 4) = 0

x + 5 = 0 or x – 4 = 0

x = – 5 or x = 4

The roots of the quadratic equation is x = – 5 or x = 4.

(iii) Find the 19^{th} term of the following A.P.:

7, 13, 19, 25,……

Solution:

7, 13, 19, 25, …

a = 7, d = 13 – 7 = 6, n = 19, t_{19} = ?

t_{n} = a + (n – 1 )d

t_{19} = 7 + (19 – 1)(6)

t_{19} = 7 + 18 × 6

t_{19} = 7 + 108

t_{19} = 115

∴ The 19^{th} term of an A.P. is 115.

(iv) A card is drawn from well shuffled pack of 52 playing cards. Find the probability that the card drawn is a face card.

Solution:

There are 52 cards.

n(S) = 52

Let A’ be the event of getting face cards.

Event A : Getting face cards. There are 12 face cards in the pack of playing cards.

n(A) = 12

P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{52}\)

P(A) = \(\frac{3}{13}\).

(v) The following table shown classification of number of workers and number of hours they work in software company. Prepare less than upper limit type cumulative frequency distribution table:

Number of hours daily | 8-10 | 10-12 | 12-14 | 14-16 |

Number of workers | 150 | 500 | 300 | 50 |

Solution:

Number of hours daily | Number of workers | Cumulative frequency |

8-10 | 150 | 150 |

10-12 | 500 | 150 + 500 = 650 |

12-14 | 300 | 650 + 300 = 950 |

14-16 | 50 | 950 + 50 = 1000 |

Question 3.

(A) Complete the following activity and rewrite it (any one): [3]

(i) The following frequency distribution table showns the classification of the number of vehicles and the volume of petrol filled in them. To find the mode of the volume of petrol filled, complete the following activity:

Class (Petrol filled in Liters) | 0.5 – 3.5 | 3.5 – 6.5 | 6.5 – 9.5 | 9.5 -12.5 | 12.5 -15.5 |

Frequency (Number of Vehicles) | 33 | 40 | 27 | 18 | 12 |

Activity:

From the given table,

Solution:

Activity:

From the given table

Modal class = 3.5 – 6.5 (∴ This class has max. frequency)

l = Lower class boundary of the modal class

f_{1} = 40 (Frequency of modal class)

f_{2} = 27 (Frequency of the class succeding the modal class)

f_{0} = 33 (frequency of the class preceding the modal class)

Mode = 3.5 + \(\left[\frac{40-33}{2(40)-33-27}\right]\) × __3__

Mode = 3.5 + \(\left[\frac{7}{80-60}\right]\) × 3

Mode = 4.55

The mode of the volume of petrol filled is 4.55.

(ii) The total value (with GST) of remote controlled toy car is ₹2360. Rate of GST is 18% on toys. Complete the following activity to find the taxable value for the toy car:

Activity:

Total value for toy car with GST = ₹2360.

Rate of GST = 18%

Let taxable value for toy car be ₹x.

∴ GST = \(\frac{18}{100}\) × x

Solution:

Activity:

Total value for toy car with GST = ₹2360.

Rate of GST 18%

Let taxable value for toy car be ₹x.

∴ GST = \(\frac{18}{100}\) × x

∴ Total value for toy car Taxable value for toy car + __GST__ [Formula]

∴ Taxable value for toy is ₹ 20,00

(B) Solve the following subquestiois (any two):

(i) Solve the following quadratic equation by formula method:

3m^{2} – m – 10 = 0

Solution:

3m^{2} – m – 10 = 0

Comparing with standard form

ax^{2} + bx + c = 0

a = 3, b = -1, c = -10

D = b^{2} – 4ac

= (-1)^{2} – 4(3)(-10)

= 1 – 12 × -10

= 1 + 120

= 121 > 0

By formula method,

The roots of the quadratic equation are m = 2 or m = \(\frac{-5}{3}\)

(ii) Solve the following simultaneous equations using Cramer’s rule:

3x – 4y = 10, 4x + 3y = 5

Solution:

3x – 4y = 10 …… (1)

4x + 3y = 5 ….. (2)

a_{1} = 3, a_{2} = 4, b_{1} = -4, b_{2} = 3, c_{1} = 10, c_{2} = 5

D = \(\left|\begin{array}{ll}

a_1 & b_1 \\

a_2 & b_2

\end{array}\right|\)

= \(\left|\begin{array}{rr}

3 & -4 \\

4 & 3

\end{array}\right|\)

= 3 × 3 – (-4 × 4)

= 9 + 16

= 25

D_{x} = \(\left|\begin{array}{ll}

c_1 & b_1 \\

c_2 & b_2

\end{array}\right|\)

= \(\left|\begin{array}{rr}

10 & -4 \\

5 & 3

\end{array}\right|\)

= 10 × 3 – (-4 × 5)

= 30 + 20

= 50

D_{y} = \(\left|\begin{array}{ll}

a_1 & c_1 \\

a_2 & c_2

\end{array}\right|\)

= \(\left|\begin{array}{cc}

3 & 10 \\

4 & 5

\end{array}\right|\)

= 3 × 5 – 4 × 10

= 15 – 40

= -25

By Cramer’s rule,

x = \(\frac{\mathrm{D} x}{\mathrm{D}}\) = \(\frac{50}{25}\) = 2

y = \(\frac{\mathrm{Dy}}{\mathrm{D}}\) = \(\frac{-25}{25}\)

= -1

The solution of the simultaneous equation is (x, y) = (2, -1).

(iii) 50 shares of face value ₹10 were purchased for market value of ₹25. Company declared 30% dividend on the shares, then find :

(1) Sum invested

(2) Dividend received

(3) Rate of return.

Solution:

No. of shares = 50

FV = ₹10

MV = ₹25

Rate of dividend = 30%

(1) Sum invested = No. of shares × MV

= 50 × 25

= ₹1250.

(2) Dividend received,

Dividend \(=\frac{\text { Rate of dividend }}{100}\) × FV

= \(\frac{30}{100}\) × 10

= ₹3

Total dividend = No. of shares × Dividend

= 50 × 3

= ₹150

Dividend received = ₹150.

(3) Rate of return,

Rate of return \(=\frac{\text { Dividend }}{\text { Sum invested }}\) × 100

= \(\frac{150}{1250}\) × 100

= \(\frac{1500}{125}\)

= 12%

Rate of dividend = 12%

(iv) One coin and a die are thrown simultaneously. Find the probability of the following events:

Event A : To get a head and a prime number.

Event B : To get a tail and an odd number.

Solution:

One coin and one die are rolled.

S = {T_{1}, T_{2}, T_{3}, T_{4}, T_{5}, T_{6}, H_{1}, H_{2}, H_{3}, H_{4}, H_{5}, H_{6}} n(S) = 12

Event A : To get a head and a prime number.

A = {H_{2}, H_{3}, H_{5}}

n(A) = 3

P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Event B: To get a tail and an odd number.

B = {T_{1}, T_{3}, T_{5})

n(B) = 3

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

P(A) = \(\frac{1}{4}\)

P(B) = \(\frac{1}{4}\)

Question 4.

Solve the following subquestions (any two): [8]

(i) A tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.

Solution:

Let bigger tap takes x hours to fill the tank and smaller tank takes x + 5 hours to fill the tank.

In one hour, Bigger tap takes \(\frac{1}{x}\) hours and smaller tap will take \(\frac{1}{x+5}\) hours and both the tap will take \(\frac{1}{6}\) hours to fill the tank.

According to the given condition,

\(\frac{1}{x}\) + \(\frac{1}{x+5}\) = \(\frac{1}{6}\)

\(\frac{x+5+x}{x(x+5)}\) = \(\frac{1}{6}\)

\(\frac{2 x+5}{x^2+5 x}\) = \(\frac{1}{6}\)

6(2x + 5) = x^{2} + 5x

12x + 30 = x^{2} + 5x

x^{2} + 5x – 12x – 30 = 0

x^{2} – 7x – 30 = 0

x^{2} – 10x + 3x – 30 = 0

x(x – 10) + 3(x – 10) = 0

(x -10) (x + 3) = 0

x – 10 = 0 or (x + 3) = 0

x = 10 or x = – 3

But x ≠ -3

∴ Time cannot be negative.

Bigger tap, x = 10 hours

Smaller tap, x + 5 = 10 + 5 = 15 hours

Bigger tap takes 10 hours and smaller tap takes 15 hours to fill the tank.

(ii) The following table shows the classification of percentage of marks of students and the number of students. Draw frequency polygon from the table without drawing histogram:

Solution:

Result (Percentage) | Class Mark | No. of Students | Points |

20-40 | 30 | 25 | (30, 25) |

40-60 | 50 | 65 | (50, 65) |

60-80 | 70 | 80 | (70, 80) |

80 -100 | 90 | 15 | (90,15) |

(iii) In a ‘Mahila Bachat Gat’ Kavita invested from the first day of month ₹20 on first day, ₹40 on second day and ₹60 on third day. If she saves like this, then what would be her total saving in the month of February 2020 ?

Solution:

2020 is a leap year.

In February month there are 29 days.

n = 29

On first day of the month Kavita invested ₹20.

On second day she invested = ₹40.

On third day she invested = ₹60.

Kavita’s investment is in A.P. 20, 40, 60, …

a = 20, n = 29, d = 20

S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

S_{29} = \(\frac{29}{2}\)[2(20) + (29 – 1)20]

= \(\frac{29}{2}\)[40 + 28 × 20]

= \(\frac{29}{2}\)[40 + 560] = \(\frac{29}{2}\) × 600

= 29 × 300 = 8700

Kavita’s total saving of the month of February is ₹8,700.

Question 5.

Solve the following subquestions (any one): [3]

(i) In the given figure, the pie diagram represents the amount spent on different sports by a school administration in a year. If the money spent on football is ₹9,000 answer the following questions:

(a) What is the total amount spent on sports ?

(b) What is the amount spent on cricket ?

Solution:

(a) Let total amount spent on sports be x.

Central angle for football = 45°

Amount spent on football = 9,000

Central angle \(=\frac{\text { Amount spent on football } \times 360}{\text { Total amount spent on sports }}\)

45 = \(\frac{9,000}{x}\) × 360

x = \(\frac{9,000 \times 360}{45}\)

x = 200 × 360

x = 72,000

Total amount spent on sports is 72,000.

(b) Let a’mount spent on cricket by y.

Central angle for cricket = 160°

Total amount spent on sports ₹72,000

Central angle \(=\frac{\text { Amount spent on cricket }}{\text { Total amount spent on sports }}\) × 360

160 = \(\frac{y}{72,000}\) × 360

y = \(\frac{160 \times 72,000}{360}\)

y = ₹32,000

Amount spent on cricket is ₹32,000.

(ii) Draw the graph of the equation x + y =4 and answer the following questions:

(a) Which type of triangle is formed by the line with X and Y-axes based on its sides.

(b) Find the area of that triangle.

Solution:

x + y = 4

x = 4 – y

Sub y = 2

x = 4 – 2

x = 2

Sub y = 0

x = 4 – 0

x = 4

Sub y = 1

x = 4 – 1

x = 3

Sub y = 3

x = 4 – 3

x = 1

x | 2 | 3 | 1 | 4 |

y | 2 | 1 | 3 | 0 |

(x, y) | (2,2) | (3,1) | (1,3) | (4,0) |

(a) Isosceles right angled triangle is formed by the line with X and Y-axes based on its sides.

(b) Height = 4 cm

Base = 4 cm

Area of triangle = \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\) × 4 × 4

= \(\frac{1}{2}\) × 16 = 8 cm^{2}

Area of triangle = 8 cm^{2}.